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Audition-Open Lines Program, Tuesday, 10-14-14 October 15, 2014

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Audition-Open Lines Program, Tuesday, 10-14-14


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Guests:  Open Lines/Auditions with Dr. David Livingston.  Topics:  Nuclear rocket propulsion was the main discussion topic.  Please direct all comments and questions regarding Space Show programs/guest(s) to the Space Show blog, https://thespaceshow.wordpress.com.  Comments and questions should be relevant to the specific Space Show program. Written Transcripts of Space Show programs are a violation of our copyright and are not permitted without prior written consent, even if for your own use. We do not permit the commercial use of Space Show programs or any part thereof, nor do we permit editing, YouTube clips, or clips placed on other private channels & websites. Space Show programs can be quoted, but the quote must be cited or referenced using the proper citation format. Contact The Space Show for further information. In addition, please remember that your Amazon purchases can help support The Space Show/OGLF. See http://www.onegiantleapfoundation.org/amazon.htm.  For those listening to archives using live365.com and rating the programs, please email me as to why you assign a specific rating to the show. This will help me bring better programming to the audience.

Welcome to this 1 hour 53 minute program focused on auditions for possible new Space Show guests plus Open Lines calls from listeners.  While there were no audition calls, two callers checked in to talk about nuclear propulsion, reactors, fuels, ISP, etc.  Both calls were interesting and instructive.  I also promoted the upcoming Next Giant Leap Conference in Hawaii from Nov. 9-13.  For more information about this conference, visit http://2014giantleap.aerospacehawaii.info.  During the first segment, I went over the audition ground rules for the show and discussed the best way for self-invited guests to be on The Space Show.  The best way is to call an Open Lines program and let us know what you have to say and why you believe you should be on the program as a guest.  Long time Space Show guest and caller Dave Ketchledge called to talk about nuclear rockets and reactors.  We covered lots of information including what it takes to actually turn on and start up a nuclear reactor.  We talked core issues & I inquired about the possibility of a gas core.  Dave talked about new and exotic nuclear fuels, the nuclear waste problem, including how the waste is treated by the advanced design Westinghouse and GE reactors plus those in France.  I asked Dave for his thoughts on the timeline for a nuclear rocket. He said probably 20 years but were we to have leadership in the area and funding, perhaps as early as ten years.  Toward the end of his call, he shared with us a few horror stories where nuclear workers had done really stupid things.

In the second segment, again there were no audition callers so I talked about how to be a Space Show guest, especially if you are self-inviting yourself to the show.  As I said before, the best way to do it is to call an Open Lines show or come in recommended by a previous guest or somebody I know.  I also spoke about listener participation through emails and phone calls with live shows rather than emailing me complaints about the program or guest after listening to the live show.  The bottom line is that we want your feedback and exchange with the guests, not after the fact.  SLS John called to talk about the Classroom show from Sunday, specifically the nuclear component of the discussion revolving around the Aquarius project proposed by Dr. Jim Logan and Dan Adamo.  John was again questioning the 900 seconds of ISP with water as the fuel.  I suggested he post his comments on the blog for that show as I was not the one to respond since Aquarius was not my project.  I did ask John to explain some basics about nuclear rockets, especially the part about the dissociation of the hydrogen atom.  We went over T/M, chemical fuel bonding, molecular weight and what happens with the very high temperatures involved in the nuclear rocket.  I also asked John for his background in the subject and for his bio as he is part of the Great SLS Webinar Debate upcoming on The Space Show on Sunday, Nov. 16 with Rick Boozer.  Finally, during this segment I highlighted some of the upcoming Space Show programs on the schedule.

Please post comments/questions on The Space Show blog.  You can reach either of our callers through me.


1. Matt - October 16, 2014

Hi John,

I think the simple Isp-formula, which uses the T/M ratio as a paramter does not apply in our case, because the average molecular mass is not constant during expansion process due recombination of dissociated radicals in the nozzle. Furthermore, significant heat is added during expansion process due to recombination. Your simple Isp formula is only valid for a heated gas, which is adiabatic expanded in a nozzle without chemical reactions during expansion, with a constant ratio isentropic exponent and without heat addition in nozzle.

If the formula would apply (what is not the case) you should also calculate the Isp using the average molecular mass of the gas species that you regarded (there are three molecules/moles:one moles oxygen atoms, two moles hydrogen atoms,weighting together 18 grams, therefore M= 6 g/mol).

Here is a link, that shows required temperature to achieve water dissocication (figure 3):


2. John (in Fort Worth) - October 15, 2014

Note: I’ve also posted this on the Sunday show as well.

The question that I had about the Isp = 900 sec at 3000 degrees C with water as the propellant is the following. In James Dewar’s book “To the End of the Solar System: The Story of the Nuclear Rocket” an Isp of 1000 sec is cited at 3000 degrees C for H2 is given. Using the sqrt( T/M) rule where T is temperature in Kelvin and M is the molar mass were can estimate Isp for other propellants. H2O has a mass of 18 in atomic mass units. So scaling is the rule which is Isp (water) = Isp (H2) x sqrt(2/18) = 1000 sec x 1/3 = 333 seconds. This assumes no disassociation of the water and puts water as propellant in the same Isp class as chemical rockets.

If one goes to the best case for Aquarius which is 100% disassociation then the result is two monotonic H particles mass = 1 and one monotonic O particle mass =16 produced for every molecule of H2O. Using the scaling approach above was have
Isp (H) = Isp (H2) x sqrt (mass of H2 / mass of H) = 1000 sec x sqrt (2 / 1) = 1000 sec x sqrt (2) = 1000 sec x 1.414 = 1,414 sec.

Similarly, the Isp (O) = Isp (H2) x sqrt (mass of H2 / mass of O) = 1000 sec x sqrt (2/16) = 354 sec. Here is the part I’m not sure about but I assume that the specific impulse of the disassociated H, H, and O is the mass weighted average of there specific impulses. There for Isp (Aquarius) = 2/18 * 1414 sec + 16/18 * 354 sec = 472 sec. So this leaves Aquarius at a specific impulse that is only slightly better than the best chemical system but only slightly more than half of the proposed 900 sec.

I’m not sure that this is the correct way to approach this problem as there can be some complexities that this basic weighted average approach misses. On the other hand I have made a favorable assumption for Aquarius in the I’ve assumed complete disassociation of water at 3000 degrees C or 3273 Kelvin.

Kirk - October 26, 2014

Hi John, are you still tracking this?

I concur with your mass weighted average of the specific impulses, although I agree that there is a danger in just assuming, without proof, that this is the proper way to do it, so let’s derive it from first principles.

Specific impulse (ISP) is defined as the force of thrust (F) divided by rate of consumption of propellant. If we use the mass rate of consumption (m-dot) we get specific impulse in units of velocity, but by using weight instead (m-dot g) we get units of time, which leads to intuitive uses when computing accelerations in units of g. So:

ISP = F / (m-dot g)

If the exhaust consists of a single species of gas, the thrust is equal to the velocity of the exhaust gas (V) times its mass rate of production (m-dot):

F = V m-dot

Substituting this into the first equation, we see that ISP is proportional to the velocity of the exhaust gas:

ISP = V / g

This is the point where we would usually bring up the kinetic theory of gases to point out that the velocity, and thus also the ISP, will be proportional sqrt(T/M) where T is the temperature and M the molecular mass.

But instead, now consider an exhaust consisting of multiple species of gases. Let w_i be the mass fraction of gas “i” and V_i be its velocity, with ISP_i = V_i / g. Sum(w_i) = 1. The rate of production of gas “i” will be m-dot_i = w_i m-dot.

The total thrust, F, will be equal to the sum of the thrust of the individual gases. That is:

F = Sum(F_i)


F_i = V_i m-dot_i = V_i w_i m-dot


ISP = F / (m-dot g)
= Sum(F_i) / (m-dot g)
= Sum(V_i w_i m-dot) / (m-dot g)
= Sum((V_i / g) w_i)
= Sum(ISP_i w_i)

which is the mass weighted average of the ISP of the individual gas species.

And I agree with your numbers, that if a given temperature will yield an ISP of 1000 sec for molecular H2, it would be 1414 sec for monatomic H, 333 sec for steam and 471 for fully disassociated H2O.

I can’t imagine where they are getting an ISP of 1000 seconds for Project Aquarius.

Kirk - October 26, 2014

Interestingly, complete disassociation of both H2 and H2O boost ISP by a factor of sqrt(2), a 41% increase. This is just a numerical coincidence caused by the fact that (2/18)sqrt(18/1) + (16/18)sqrt(18/16) simplifies to sqrt(2).

So for a given temperature:
ISP_Disassociated_Hydrogen = sqrt(2) ISP_H2 = 1.41 ISP_H2
ISP_Steam = (1/3) ISP_H2 = 0.33 ISP_H2
ISP_Disassociated_Water = (1/3)sqrt(2) ISP_H2 = 0.74 ISP_H2

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